
Algebra PreAlgebra and Basic Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
November 6th, 2017, 10:37 AM  #1 
Newbie Joined: Nov 2017 From: Europe Posts: 2 Thanks: 0  error: linear inequality (absolute function)
hello, I tried to solve the following inequality: abs(2x+1)>2/3x+1 but get wrong results: 2x+1>2/3x1 I add 1 then +2x then multiply with 3 and divide by 4 I get: x<3/2 and 2x+1>2/3x+1 cancel out 1 then +2x 0>8/3x so I get x<0 surely this inequality is true for some x>0  do you spot the error? best, richard Last edited by skipjack; November 7th, 2017 at 12:30 AM. 
November 6th, 2017, 07:01 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,642 Thanks: 960 Math Focus: Elementary mathematics and beyond 
There are two cases; one where 2x + 1 is positive and one where it's negative. It's zero at x = 1/2. Does that help? Hint: make a sign chart  if you know how. If not, consult Google (or a search engine of your choice). You could also graph both functions on the same axes. Last edited by greg1313; November 6th, 2017 at 07:10 PM. 
November 7th, 2017, 05:07 AM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,824 Thanks: 752 
Is the right side "(2/3)x+ 1" or "2/(3x)+ 1" or "2/(3x+ 1)"? Please use parentheses!

November 8th, 2017, 08:56 AM  #4 
Newbie Joined: Nov 2017 From: Europe Posts: 2 Thanks: 0 
Hi Country Boy, sorry that's the one I meant: abs(2x+1)>(2/3)x+1 Last edited by skipjack; November 8th, 2017 at 11:04 AM. 
November 8th, 2017, 11:20 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 18,155 Thanks: 1422 
Let's first consider x < 1/2, so that abs(2x + 1) = 2x + 1. To solve 2x + 1 > (2/3)x + 1, add 2x  1 to both sides to get 0 > (8/3)x, which is equivalent to x < 0. So far, this matches what you did. Now let's consider x $\small\geqslant$ 1/2, so that abs(2x + 1) = 2x  1. To solve 2x  1 > (2/3)x + 1, subtract (2/3)x  1 from both sides to get (4/3)x > 2, which is equivalent to x > 3/2. 
November 10th, 2017, 04:09 AM  #6 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,824 Thanks: 752 
By the way this is NOT a "linear" inequality. Absolute value is not a linear function.


Tags 
absolute, error, function, inequality, linear 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Maximum absolute error when calculating a difference using two rounded down values.  trampolinerist  Algebra  1  October 29th, 2017 07:16 PM 
Triangle Inequality: Prove Absolute Value Inequality  StillAlive  Calculus  5  September 3rd, 2016 12:45 AM 
Absolute Value Inequality with Absolute Values on Both Sides  shiseonji  Algebra  2  September 24th, 2013 09:36 AM 
absolute value inequality  ommmmid  Algebra  5  March 19th, 2013 10:38 PM 
least absolute error minimization with quadratic constraint  benoit  Linear Algebra  0  December 1st, 2010 03:34 AM 